a
In \(\mathrm{BF}_{3}\), the ground state electronic configuration of Boron is \(1 \mathrm{s}^{2} 2 \mathrm{s}^{2} 2 \mathrm{P} \mathrm{x}^{1}\). excited state electronic configuration is \(1 \mathrm{s}^{2} 2 \mathrm{s}^{1} 2 \mathrm{P} \mathrm{x}^{1} 2 \mathrm{Py}^{1}\)

One electron from one 's' orbital and two electrons from 'p' orbital overlaps with 'p' orbital electrons of Flourine to form three \(\mathrm{sp}^{2}\) hybridised bonds. So, \(\mathrm{BF}_{3}\) has sp \(^{2}\) hybridisation.

Flourine (a) But in \(\mathrm{H}_{3} \mathrm{O}^{+}, \mathrm{NH}_{3}\) and \(\mathrm{P} \mathrm{Cl}_{3}\), the hybridisation involves one electron from 's \(^{\prime}\) orbital, 3 electrons for \({ }^{\prime} \mathrm{p}^{\prime}\) orbitals in central orbital forming \(\mathrm{sp}^{3}\) hybridisation.